Stuck on one of the substitution method steps

[Jeff12341234]

I put it in std form, did the homogeneous test. it passed with degree 2. I substituted y=ux and dy=udx+xdu and now I'm stuck. it needs to be simplified somehow but I don't know if ux is one var or if it's u*x. Same goes for udx and xdu. Is it really u*dx+x*du? even assuming that is correct, it doesn't get any simpler. What does the next step look like for this problem?

 

[lurflurf]

That is good so far, the substitution has transformed the equation into separable form, divide to complete separation.

$$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$$$\frac{((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})}{x^3(3u^2-2)}=\frac{\mathrm{dx}}{x}+\frac{2u\mathrm{du}}{3u^2-2}=0$$

 

[Jeff12341234]

Where did you get the denominator x³(3u²-2) from?

What does f(u), g(x), p(u), and q(x) each equal?

 

[lurflurf]

In this example
$$\mathrm{f}(u)=3u^2-2 \\ \mathrm{g}(x)=x^2 \\ \mathrm{p}(u)=2u \\ \mathrm{q}(x)=x^3$$

This is just separation of variables
$$((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})=(3u^2-2)(x^2)\mathrm{dx}+(2u)(x^3)\mathrm{du}$$
So we divide by the the factors that do not match their differentials.

 

[Jeff12341234]

$$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$

Is the above eq some sort of shortcut that is explained somewhere. I've never seen it in all of the lessons I've read.

I understand how I got to here:
$$((ux)^2-2x^2)\mathrm{dx}+2x(ux)(x \mathrm{du}+u\mathrm{dx})$$

and how you got from here:
$$(3u^2-2)(x^2)\mathrm{dx}+(2u)(x^3)\mathrm{du}$$ to the next part but not the step(s) in between. I tried working it out several times but could never separate the variables completely.

 

[lurflurf]

$$\frac{\mathrm{f}(u)\mathrm{g}(x) \mathrm{dx}+\mathrm{p}(u) \mathrm{q}(x)\mathrm{du}}{\mathrm{f}(u) \mathrm{q}(x)}=\frac{\mathrm{g}(x)}{\mathrm{q}(x)}\mathrm{dx}+\frac{\mathrm{p}(u) }{\mathrm{f}(u)}\mathrm{du}$$
That is just the separation of variables equation.
Write the equation in the form of each differential multiplied by a function of each variable, divide by each non-matching function to give each differential multiplied by a matching function.


$$((ux)^2-2x^2) \, \mathrm{dx}+2x(ux)(x \, \mathrm{du}+u \, \mathrm{dx})= x^2(u^2 -2) \, \mathrm{dx}+2x^2 u(x \, \mathrm{du}+u \, \mathrm{dx})\\ =x^2(u^2 -2) \, \mathrm{dx}+(2x^2 u)x \, \mathrm{du}+(2x^2 u)u \,\mathrm{dx}\\ =x^2(u^2 -2+2u^2) \, \mathrm{dx}+2x^3 u \, \mathrm{du} \\ =(3u^2-2)(x^2) \, \mathrm{dx}+(2u)(x^3) \, \mathrm{du}$$

 

[Jeff12341234]

thanks. I got it now.